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a particle with certain initial velocity is uniformly accelerated . if it covers 25 m in 5th second and 49m in 7th second of it's motion determine the distance travelled by it in 10sec. |
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Answer» S = v t + 1/2 a t^2S(5) - S(4) = v (5 - 4) + 1/2 a (25 - 16) TRAVELED in 5th secS(7) - S(6) = v (7 - 6) + 1/2 a (49 - 36) " " 7TH "25 = v + a/2 * 949 = v + a/2 * 1324 = 2 a SUBTRACTING equationsa = 122 v = 74 - 6 * 22 adding equationsv = -29S(10) = -29 * 10 + 12 * 100 / 2 = 310 m total DISTANCE in 10 secCheck:S(5) - S(4) = -29 * (5 - 4) + 1/2 * 12 (25 - 16) = 54 - 29 = 25 |
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